On series of functions with the Baire property

Abstract

Hugo Steinhaus in 1928 proved three theorems concerning convergence almost everywhere and divergence of series of measurable functions. We investigate similarities and differences in the behaviour of sequences of functions with the Baire property for convergence everywhere except a set of the first category.

Recall three classical theorems concerning series of real numbers.

FormalPara Theorem 1

(Abel and Dini, [2, Chapter IX, Section 39, (173)]) If , $a_k\geqslant 0$ for $k\in \mathbb {N}$, then there exists a sequence of positive numbers such that and .

FormalPara Theorem 2

(Dini, [2, Chapter IX, Section 39, (175), 4 and Section 41, (178), A]) If , $a_k\geqslant 0$ for $k\in \mathbb {N}$, then there exists a sequence of positive numbers such that and .

FormalPara Theorem 3

(Stieltjes, [2, Chapter IX, Section 41, (178), I]) For each sequence of positive numbers such that there exists a series , $a_k>0$ for $k\in \mathbb {N}$, such that and .

These theorems show that there does not exist neither a series which converges slower than any other series nor a series which diverges slower than any other series. This implies among others that no comparison test can be effective with all series.

Theorem 1 remains true also for a sequence $\{S_i\}_{i\in \mathbb {N}}$ of divergent series of positive numbers. Namely, there exists a sequence independent from i such that all series , $i\in \mathbb {N}$, are divergent. It was proved in [4, 5] with the use of the Banach–Steinhaus theorem. In the same paper Hugo Steinhaus dealt with the following question of Stanisław Ruziewicz: Suppose that a series of positive functions , for $k\in \mathbb {N}$, is divergent for each . Is it possible to find a sequence of positive numbers such that and for each t? The answer is negative, there exists a sequence of measurable functions such that for each sequence there exists a point such that .

Let ${\varvec{\Lambda }}$ be the set of all positive sequences convergent to 0 ($\mathrm{card}\,{\varvec{\Lambda }} = \mathfrak {C}$). Take a set , $m(E) =0$, $\mathrm{card}\,E =\mathfrak {C}$ (here stands for the Lebesgue measure). There exists a function $t:{\varvec{\Lambda }} \xrightarrow {1-1\,} E$ (denotations of Steinhaus [4, 5]). For each there exists such that , $a_k^\Lambda >0$, and (again Stieltjes). Put for $t = t(\Lambda )$ and for $t\notin E$, $k\in \mathbb {N}$. All functions $a_k$ are measurable ($= {1}/{k}$ a.e.) and

If one requires only divergence or convergence almost everywhere, the corresponding versions of Theorems 1, 2, 3 remain true. Three theorems below are also due to Steinhaus [4, 5].

FormalPara Theorem 4

If is a sequence of measurable functions, for $k\in \mathbb {N}$ and almost everywhere on , then there exists a sequence of positive numbers such that and for almost every .

FormalPara Theorem 5

If is a sequence of measurable functions, for $k\in \mathbb {N}$ and for almost every , then there exists a sequence of positive numbers such that and for almost every .

FormalPara Theorem 6

If is a sequence of measurable functions, for $k\in \mathbb {N}$ and almost everywhere on , then there exists a sequence of positive numbers such that and almost everywhere.

Below we shall prove that Theorem 4 has a satisfactory analogue for the Baire property while Theorems 5 and 6 do not.

FormalPara Theorem 7

If is a sequence of functions with the Baire property, for $k\in \mathbb {N}$ and everywhere except a set of the first category on , then there exists a sequence of positive numbers such that and everywhere except a set of the first category.

FormalPara Proof

We shall need a simple lemma:

FormalPara Lemma

If $\{f_n\}_{n\in \mathbb N}$ is an increasing sequence of functions with the Baire property, for $n\in \mathbb {N}$ and everywhere except a set of the first category, then for each $p >0$ and for each interval there exists such that for $n\geqslant n_0$ is of the second category (so $E_n$ is residual in some subinterval as a set having the Baire property).

FormalPara Proof of Lemma

Let for $n\in \mathbb {N}$. Since is residual in , there exists such that $E_{n_0}$ is of the second category, each $E_n$ for $n\geqslant n_0$ is also of the second category because the sequence $\{E_n\}_{n\in \mathbb {N}}$ is ascending.$\quad \square$

Now, the existence of an increasing sequence of natural numbers will follow from the lemma. Let $n_1$ be a natural number for which the set is of the second category. Let $n_2 >n_1$ be a natural number for which both sets , are of the second category. Suppose that we have chosen $n_1< n_2<\dots <n_i$. Let $n_{i+1} >n_i$ be a natural number for which all sets

are of the second category.

Now put (similarly as in [4, 5]): $\uplambda _k =1$ for $1\leqslant k\leqslant n_1$ and $\uplambda _k ={1}/({k+1})$ for $n_i+1 \leqslant k\leqslant n_{i+1}$. We obviously have . At the same time for . Hence for . From the construction it follows that A is residual in . $\square$

To show that analogues of Theorems 5 and 6 do not hold for the Baire property we shall construct a sequence $\{f_n\}_{n\in \mathbb {N}}$ of functions with the Baire property which converges pointwise to zero and does not converge uniformly on any set of the second category with the Baire property. One can find an example of such sequence of functions in [3, Chapter 8], consisting of continuous functions, but in our construction the sequence is non-increasing, which enables us to build a series of functions with non-negative terms.

Let for $n\in \mathbb {N}$ and , , for $n\in \mathbb {N}$. Put

(1)

If $f_1, f_2,\dots , f_{n-1}$ are already defined, put

(2)

Obviously $f_{n+1}(x)\leqslant f_n(x)$ for each $n\in \mathbb {N}$ and . We shall show that for each . Clearly, if the binary expansion of x contains a finite number of 1’s (or equivalently, a finite number of 0’s), then $f_n(x) =0$ for sufficiently big $n\in \mathbb {N}$. Suppose now that the binary expansion of x contains infinitely many 0’s as well as infinitely many 1’s. If $x = (0, a_1 ,a_2,\dots , a_k,\dots )_2$ and $a_k =1$, then from the definition of $f_k$ it follows that $f_k(x) ={1}/({k+1})$; if $a_k =0$, then $f_k(x) ={1}/({j+1})$, where and $j \rightarrow \infty$ as $k\rightarrow \infty$. Hence for all .

Suppose now that is a set of the second category with the Baire property. Then there exists an interval and a set P of the first category such that . We shall show that $\{f_n\}_{n\in \mathbb {N}}$ is not uniformly convergent on . Indeed, there exists such that at least one component of $B_{n_0}$ is included in , so there exists a point and . From the construction of $\{f_n\}_{n\in \mathbb {N}}$ it follows that for each $n >n_0$ at least one component of $A_n$ is included in (a, b), so for each $n >n_0$ there exists a point such that . Hence $\{f_n\}_{n\in \mathbb {N}}$ is not uniformly convergent on and obviously on E.

FormalPara Theorem 8

There exists a sequence of functions with the Baire property, for $k\in \mathbb {N}$ and for each , such that for every sequence of positive numbers tending to infinity the series is divergent to infinity on a set residual in .

FormalPara Proof

Put and for $k\geqslant 2$, where $\{f_n\}_{n\in \mathbb {N}}$ is the sequences (1)–(2). Then for $k\in \mathbb {N}$, so everywhere except a denumerable set. Observe that for $t\in B_k$, since and on $B_k$. From the construction of $\{f_n\}_{n\in \mathbb {N}}$ it follows immediately that on for each $p\in \mathbb {N}$.

Let $\{p_k\}_{k\in \mathbb {N}}$ be an increasing sequence of natural numbers with for each $k\in \mathbb {N}$. Then we have for . Hence for and the last set is residual in . $\square$

FormalPara Theorem 9

There exists a sequence of functions with the Baire property, for $k\in \mathbb {N}$ and for , such that for each sequence of positive numbers with the series is divergent to infinity on a set residual in .

FormalPara Proof

Let be the sequence of functions (1)–(2). Suppose that . There exists such that . Observe that for $t\in (0, {1}/{2^{n_1}})$, so on this interval. There exists such that . Again observe that for $t\in (0, {1}/{2^{n_2}})\cup (1/2, 1/2 +{1}/{2^{n_2}})$ (actually the value of all functions above is equal to 1 in the first interval and to 1 / 2 in the second one), so on this set. Suppose that we have found $n_1< n_2<\cdots < n_i$. Let $n_{i+1}$ be a number such that . Similarly, we observe that on an open set $E_i$ such that the intersection of $E_i$ with each component of $A_{i+1}$ is nonempty (namely it is equal to the interval ). Hence for $t\in E_i$. Finally for and the last set is residual in . $\square$

We show another similarity and difference between measure and category. Bartle in [1] proved the following theorem:

FormalPara Theorem 10

If a sequence $\{f_n\}_{n\in \mathbb {N}}$ of measurable functions on converges almost uniformly to f, then it satisfies the vanishing restriction with respect to f. If $\{f_n\}_{n\in \mathbb {N}}$ converges in measure to f and $\{f_n\}_{n\in \mathbb {N}}$ satisfies the vanishing restriction with respect to f, then $\{f_n\}_{n\in \mathbb {N}}$ converges almost uniformly to f.

Here the convergence almost uniformly means that there exists a sequence $\{B_i\}_{i\in \mathbb {N}}$ of measurable sets such that is a nullset) and $f_{n|B_i} \rightrightarrows f_{|B_i}$ for each $i\in \mathbb {N}$ as $n\rightarrow \infty$ (cf. Egorov’s theorem).

The sequence $\{f_n\}_{n\in \mathbb {N}}$ of measurable functions satisfies the vanishing restriction with respect tof if for all $\upalpha >0$ we have

Since the sequence $\bigl \{E_n^f(\upalpha )\bigr \}{}_{n\in \mathbb {N}}$ is descending, the last condition means is a nullset.

We shall say that a sequence $\{f_n\}_{n\in \mathbb {N}}$ of functions with the Baire property converges to fB-almost uniformly if there exists a sequence $\{B_i\}_{i\in \mathbb {N}}$ of sets with the Baire property such that is of the first category and $f_{n|B_i}\rightrightarrows f_{|B_i}$ for each $i\in \mathbb {N}$ as $n\rightarrow \infty$. We can (and shall) suppose that the sequence $\{B_i\}_{i\in \mathbb {N}}$ is ascending.

We shall say that a sequence $\{f_n\}_{n\in \mathbb {N}}$ satisfies the B-vanishing restriction with respect tof if for all $\upalpha >0$ the set is of the first category.

FormalPara Theorem 11

If a sequence $\{f_n\}_{n\in \mathbb {N}}$ of functions with the Baire property on converges B-almost uniformly to f, then it satisfies the B-vanishing condition with respect to f. There exists a sequence $\{f_n\}_{n\in \mathbb {N}}$ of functions with the Baire property on [0, 1] which converges everywhere to $f\equiv 0$ and satisfies the B-vanishing restriction with respect to f but which does not converge B-almost uniformly to f.

FormalPara Proof

Suppose that $\{B_i\}_{i\in \mathbb {N}}$ is an increasing sequence of sets with the Baire property such that is of the first category and $f_{n|B_i}\rightrightarrows f_{|B_i}$ for each $i\in \mathbb {N}$ as $n\rightarrow \infty$.

Fix $\upalpha >0$. For each $i\in \mathbb {N}$ there exists such that for each $x\in B_i$ and $n\geqslant n_{\upalpha , i}$ we have $|f_n(x) - f(x)|\leqslant \upalpha$. So and which completes the proof of the first part of the theorem.

To prove the second part it is sufficient to observe that the sequence $\{f_n\}_{n\in \mathbb {N}}$ of functions (1)–(2) satisfies the B-vanishing restriction. If this sequence were convergent uniformly to the zero-function on a set with the Baire property, then E would be of the first category. Hence it is clear that $\{f_n\}_{n\in \mathbb {N}}$ does not converge B-almost uniformly to the zero-function. $\square$